Mixing
Equivalent form of the Whitehead problem
$begingroup$
Let $M$ be an $R$-module and consider the following statement.
$M$ is projective whenever the obvious group map $tau:
text{Hom}_R(M, R) to text{Hom}_R(M, {R}/{I})$ is surjective for any
ideal $Isubset R$.
I have read that when $R = mathbb{Z}$, determining the truth of this statement is equivalent to deciding the Whitehead problem, which asks whether or not the abelian group $M$ must be free so long as every group extension of $M$ by $mathbb{Z}$ is trivial.
I know that being free is equivalent to being projective for abelian groups. But why is the surjectivity of $tau$ equivalent to the condition that $text{Ext}(M, mathbb{Z}) =0$? I don't see how the fact that $tau$ is surjective implies that any such extension splits.
I may be missing something obvious, but could someone please explain?
Here is one reference for what I am talking about.
Google book page.
abstract-algebra group-theory logic homological-algebra
asked Jan 20 at 17:29
CuriousKid7CuriousKid7
1,691717
$endgroup$
add a comment |
$begingroup$
Let $M$ be an $R$-module and consider the following statement.
$M$ is projective whenever the obvious group map $tau:
text{Hom}_R(M, R) to text{Hom}_R(M, {R}/{I})$ is surjective for any
ideal $Isubset R$.
I have read that when $R = mathbb{Z}$, determining the truth of this statement is equivalent to deciding the Whitehead problem, which asks whether or not the abelian group $M$ must be free so long as every group extension of $M$ by $mathbb{Z}$ is trivial.
I know that being free is equivalent to being projective for abelian groups. But why is the surjectivity of $tau$ equivalent to the condition that $text{Ext}(M, mathbb{Z}) =0$? I don't see how the fact that $tau$ is surjective implies that any such extension splits.
I may be missing something obvious, but could someone please explain?
Here is one reference for what I am talking about.
Google book page.
abstract-algebra group-theory logic homological-algebra
asked Jan 20 at 17:29
CuriousKid7CuriousKid7
1,691717
$endgroup$
add a comment |
$begingroup$
Let $M$ be an $R$-module and consider the following statement.
$M$ is projective whenever the obvious group map $tau:
text{Hom}_R(M, R) to text{Hom}_R(M, {R}/{I})$ is surjective for any
ideal $Isubset R$.
I have read that when $R = mathbb{Z}$, determining the truth of this statement is equivalent to deciding the Whitehead problem, which asks whether or not the abelian group $M$ must be free so long as every group extension of $M$ by $mathbb{Z}$ is trivial.
I know that being free is equivalent to being projective for abelian groups. But why is the surjectivity of $tau$ equivalent to the condition that $text{Ext}(M, mathbb{Z}) =0$? I don't see how the fact that $tau$ is surjective implies that any such extension splits.
I may be missing something obvious, but could someone please explain?
Here is one reference for what I am talking about.
Google book page.
abstract-algebra group-theory logic homological-algebra
asked Jan 20 at 17:29
CuriousKid7CuriousKid7
1,691717
$endgroup$
Let $M$ be an $R$-module and consider the following statement.
$M$ is projective whenever the obvious group map $tau:
text{Hom}_R(M, R) to text{Hom}_R(M, {R}/{I})$ is surjective for any
ideal $Isubset R$.
I have read that when $R = mathbb{Z}$, determining the truth of this statement is equivalent to deciding the Whitehead problem, which asks whether or not the abelian group $M$ must be free so long as every group extension of $M$ by $mathbb{Z}$ is trivial.
I know that being free is equivalent to being projective for abelian groups. But why is the surjectivity of $tau$ equivalent to the condition that $text{Ext}(M, mathbb{Z}) =0$? I don't see how the fact that $tau$ is surjective implies that any such extension splits.
I may be missing something obvious, but could someone please explain?
Here is one reference for what I am talking about.
Google book page.
abstract-algebra group-theory logic homological-algebra
abstract-algebra group-theory logic homological-algebra
asked Jan 20 at 17:29
CuriousKid7CuriousKid7
1,691717
asked Jan 20 at 17:29
CuriousKid7CuriousKid7
1,691717
edited Jan 20 at 19:46
CuriousKid7
asked Jan 20 at 17:29
CuriousKid7CuriousKid7
1,691717
asked Jan 20 at 17:29
CuriousKid7CuriousKid7
1,691717
asked Jan 20 at 17:29
CuriousKid7CuriousKid7
1,691717
1,691717
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
This is incorrect: surjectivity of $tau$ is strictly weaker than $operatorname{Ext}(M,mathbb{Z})=0$. Specifically, surjectivity of $tau$ is instead equivalent to $operatorname{Ext}(M,mathbb{Z})$ being torsion-free. The statement that you quoted is not equivalent to Whitehead's problem for $R=mathbb{Z}$ and in fact is provably false for $R=mathbb{Z}$. For instance, if $M=mathbb{Q}$, $operatorname{Hom}(mathbb{Q},mathbb{Z}/I)=0$ for any ideal $Isubseteqmathbb{Z}$, so $tau$ is trivially always surjective, but $mathbb{Q}$ is not projective.
Here's how you prove that surjectivity of $tau$ is equivalent to $operatorname{Ext}(M,mathbb{Z})$ being torsion free. Note that for any $Isubseteq mathbb{Z}$, we have a long exact sequence $$0to operatorname{Hom}(M,I)tooperatorname{Hom}(M,mathbb{Z})stackrel{tau}tooperatorname{Hom}(M,mathbb{Z}/I)stackrel{delta}tooperatorname{Ext}(M,I)stackrel{f}tooperatorname{Ext}(M,mathbb{Z})tooperatorname{Ext}(M,mathbb{Z}/I)to 0.$$
In particular, we see that $tau$ is surjective iff $delta=0$ iff $f$ is injective. Note moreover that if $I$ is nontrivial, then $I=nmathbb{Z}$ for some $n>0$ and so we can identify $f$ with the multiplication by $n$ map $operatorname{Ext}(M,mathbb{Z})tooperatorname{Ext}(M,mathbb{Z})$. So, $tau$ is surjective for $I=nmathbb{Z}$ iff $operatorname{Ext}(M,mathbb{Z})$ has no $n$-torsion. We conclude that $tau$ is surjective for all $I$ iff $operatorname{Ext}(M,mathbb{Z})$ is torsion-free.
answered Jan 20 at 18:47
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
Thank you, but I keep reading that my quoted statement is the dual of Baer's criterion for injective modules and that this reduces to the Whitehead problem for $R= mathbb{Z}$. Do you have any idea what the correct version of this claim is supposed to look like?
$endgroup$
– CuriousKid7
Jan 20 at 19:05
$begingroup$
Where are you reading that? I don't think there's any sense in which the Whitehead problem is literally dual to Baer's criterion, though they are certainly related.
$endgroup$
– Eric Wofsey
Jan 20 at 19:35
$begingroup$
I've given the link of one reference in my OP. But you are certainly right.
$endgroup$
– CuriousKid7
Jan 20 at 19:45
2
$begingroup$
I think it's simply an error in that text where the authors were a bit over-eager to fit a remark on Whitehead's problem into their narrative.
$endgroup$
– Eric Wofsey
Jan 20 at 19:50
$begingroup$
Yes, I agree. In fact, the correct formulation is much more delicate. It seems to be known as "Faith's problem on $R$-projectivity" if you're interested.
$endgroup$
– CuriousKid7
Jan 20 at 19:52
add a comment |
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$begingroup$
This is incorrect: surjectivity of $tau$ is strictly weaker than $operatorname{Ext}(M,mathbb{Z})=0$. Specifically, surjectivity of $tau$ is instead equivalent to $operatorname{Ext}(M,mathbb{Z})$ being torsion-free. The statement that you quoted is not equivalent to Whitehead's problem for $R=mathbb{Z}$ and in fact is provably false for $R=mathbb{Z}$. For instance, if $M=mathbb{Q}$, $operatorname{Hom}(mathbb{Q},mathbb{Z}/I)=0$ for any ideal $Isubseteqmathbb{Z}$, so $tau$ is trivially always surjective, but $mathbb{Q}$ is not projective.
Here's how you prove that surjectivity of $tau$ is equivalent to $operatorname{Ext}(M,mathbb{Z})$ being torsion free. Note that for any $Isubseteq mathbb{Z}$, we have a long exact sequence $$0to operatorname{Hom}(M,I)tooperatorname{Hom}(M,mathbb{Z})stackrel{tau}tooperatorname{Hom}(M,mathbb{Z}/I)stackrel{delta}tooperatorname{Ext}(M,I)stackrel{f}tooperatorname{Ext}(M,mathbb{Z})tooperatorname{Ext}(M,mathbb{Z}/I)to 0.$$
In particular, we see that $tau$ is surjective iff $delta=0$ iff $f$ is injective. Note moreover that if $I$ is nontrivial, then $I=nmathbb{Z}$ for some $n>0$ and so we can identify $f$ with the multiplication by $n$ map $operatorname{Ext}(M,mathbb{Z})tooperatorname{Ext}(M,mathbb{Z})$. So, $tau$ is surjective for $I=nmathbb{Z}$ iff $operatorname{Ext}(M,mathbb{Z})$ has no $n$-torsion. We conclude that $tau$ is surjective for all $I$ iff $operatorname{Ext}(M,mathbb{Z})$ is torsion-free.
answered Jan 20 at 18:47
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
Thank you, but I keep reading that my quoted statement is the dual of Baer's criterion for injective modules and that this reduces to the Whitehead problem for $R= mathbb{Z}$. Do you have any idea what the correct version of this claim is supposed to look like?
$endgroup$
– CuriousKid7
Jan 20 at 19:05
$begingroup$
Where are you reading that? I don't think there's any sense in which the Whitehead problem is literally dual to Baer's criterion, though they are certainly related.
$endgroup$
– Eric Wofsey
Jan 20 at 19:35
$begingroup$
I've given the link of one reference in my OP. But you are certainly right.
$endgroup$
– CuriousKid7
Jan 20 at 19:45
2
$begingroup$
I think it's simply an error in that text where the authors were a bit over-eager to fit a remark on Whitehead's problem into their narrative.
$endgroup$
– Eric Wofsey
Jan 20 at 19:50
$begingroup$
Yes, I agree. In fact, the correct formulation is much more delicate. It seems to be known as "Faith's problem on $R$-projectivity" if you're interested.
$endgroup$
– CuriousKid7
Jan 20 at 19:52
add a comment |
$begingroup$
This is incorrect: surjectivity of $tau$ is strictly weaker than $operatorname{Ext}(M,mathbb{Z})=0$. Specifically, surjectivity of $tau$ is instead equivalent to $operatorname{Ext}(M,mathbb{Z})$ being torsion-free. The statement that you quoted is not equivalent to Whitehead's problem for $R=mathbb{Z}$ and in fact is provably false for $R=mathbb{Z}$. For instance, if $M=mathbb{Q}$, $operatorname{Hom}(mathbb{Q},mathbb{Z}/I)=0$ for any ideal $Isubseteqmathbb{Z}$, so $tau$ is trivially always surjective, but $mathbb{Q}$ is not projective.
Here's how you prove that surjectivity of $tau$ is equivalent to $operatorname{Ext}(M,mathbb{Z})$ being torsion free. Note that for any $Isubseteq mathbb{Z}$, we have a long exact sequence $$0to operatorname{Hom}(M,I)tooperatorname{Hom}(M,mathbb{Z})stackrel{tau}tooperatorname{Hom}(M,mathbb{Z}/I)stackrel{delta}tooperatorname{Ext}(M,I)stackrel{f}tooperatorname{Ext}(M,mathbb{Z})tooperatorname{Ext}(M,mathbb{Z}/I)to 0.$$
In particular, we see that $tau$ is surjective iff $delta=0$ iff $f$ is injective. Note moreover that if $I$ is nontrivial, then $I=nmathbb{Z}$ for some $n>0$ and so we can identify $f$ with the multiplication by $n$ map $operatorname{Ext}(M,mathbb{Z})tooperatorname{Ext}(M,mathbb{Z})$. So, $tau$ is surjective for $I=nmathbb{Z}$ iff $operatorname{Ext}(M,mathbb{Z})$ has no $n$-torsion. We conclude that $tau$ is surjective for all $I$ iff $operatorname{Ext}(M,mathbb{Z})$ is torsion-free.
answered Jan 20 at 18:47
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
Thank you, but I keep reading that my quoted statement is the dual of Baer's criterion for injective modules and that this reduces to the Whitehead problem for $R= mathbb{Z}$. Do you have any idea what the correct version of this claim is supposed to look like?
$endgroup$
– CuriousKid7
Jan 20 at 19:05
$begingroup$
Where are you reading that? I don't think there's any sense in which the Whitehead problem is literally dual to Baer's criterion, though they are certainly related.
$endgroup$
– Eric Wofsey
Jan 20 at 19:35
$begingroup$
I've given the link of one reference in my OP. But you are certainly right.
$endgroup$
– CuriousKid7
Jan 20 at 19:45
2
$begingroup$
I think it's simply an error in that text where the authors were a bit over-eager to fit a remark on Whitehead's problem into their narrative.
$endgroup$
– Eric Wofsey
Jan 20 at 19:50
$begingroup$
Yes, I agree. In fact, the correct formulation is much more delicate. It seems to be known as "Faith's problem on $R$-projectivity" if you're interested.
$endgroup$
– CuriousKid7
Jan 20 at 19:52
add a comment |
$begingroup$
This is incorrect: surjectivity of $tau$ is strictly weaker than $operatorname{Ext}(M,mathbb{Z})=0$. Specifically, surjectivity of $tau$ is instead equivalent to $operatorname{Ext}(M,mathbb{Z})$ being torsion-free. The statement that you quoted is not equivalent to Whitehead's problem for $R=mathbb{Z}$ and in fact is provably false for $R=mathbb{Z}$. For instance, if $M=mathbb{Q}$, $operatorname{Hom}(mathbb{Q},mathbb{Z}/I)=0$ for any ideal $Isubseteqmathbb{Z}$, so $tau$ is trivially always surjective, but $mathbb{Q}$ is not projective.
Here's how you prove that surjectivity of $tau$ is equivalent to $operatorname{Ext}(M,mathbb{Z})$ being torsion free. Note that for any $Isubseteq mathbb{Z}$, we have a long exact sequence $$0to operatorname{Hom}(M,I)tooperatorname{Hom}(M,mathbb{Z})stackrel{tau}tooperatorname{Hom}(M,mathbb{Z}/I)stackrel{delta}tooperatorname{Ext}(M,I)stackrel{f}tooperatorname{Ext}(M,mathbb{Z})tooperatorname{Ext}(M,mathbb{Z}/I)to 0.$$
In particular, we see that $tau$ is surjective iff $delta=0$ iff $f$ is injective. Note moreover that if $I$ is nontrivial, then $I=nmathbb{Z}$ for some $n>0$ and so we can identify $f$ with the multiplication by $n$ map $operatorname{Ext}(M,mathbb{Z})tooperatorname{Ext}(M,mathbb{Z})$. So, $tau$ is surjective for $I=nmathbb{Z}$ iff $operatorname{Ext}(M,mathbb{Z})$ has no $n$-torsion. We conclude that $tau$ is surjective for all $I$ iff $operatorname{Ext}(M,mathbb{Z})$ is torsion-free.
answered Jan 20 at 18:47
Eric WofseyEric Wofsey
188k14216346
$endgroup$
This is incorrect: surjectivity of $tau$ is strictly weaker than $operatorname{Ext}(M,mathbb{Z})=0$. Specifically, surjectivity of $tau$ is instead equivalent to $operatorname{Ext}(M,mathbb{Z})$ being torsion-free. The statement that you quoted is not equivalent to Whitehead's problem for $R=mathbb{Z}$ and in fact is provably false for $R=mathbb{Z}$. For instance, if $M=mathbb{Q}$, $operatorname{Hom}(mathbb{Q},mathbb{Z}/I)=0$ for any ideal $Isubseteqmathbb{Z}$, so $tau$ is trivially always surjective, but $mathbb{Q}$ is not projective.
Here's how you prove that surjectivity of $tau$ is equivalent to $operatorname{Ext}(M,mathbb{Z})$ being torsion free. Note that for any $Isubseteq mathbb{Z}$, we have a long exact sequence $$0to operatorname{Hom}(M,I)tooperatorname{Hom}(M,mathbb{Z})stackrel{tau}tooperatorname{Hom}(M,mathbb{Z}/I)stackrel{delta}tooperatorname{Ext}(M,I)stackrel{f}tooperatorname{Ext}(M,mathbb{Z})tooperatorname{Ext}(M,mathbb{Z}/I)to 0.$$
In particular, we see that $tau$ is surjective iff $delta=0$ iff $f$ is injective. Note moreover that if $I$ is nontrivial, then $I=nmathbb{Z}$ for some $n>0$ and so we can identify $f$ with the multiplication by $n$ map $operatorname{Ext}(M,mathbb{Z})tooperatorname{Ext}(M,mathbb{Z})$. So, $tau$ is surjective for $I=nmathbb{Z}$ iff $operatorname{Ext}(M,mathbb{Z})$ has no $n$-torsion. We conclude that $tau$ is surjective for all $I$ iff $operatorname{Ext}(M,mathbb{Z})$ is torsion-free.
answered Jan 20 at 18:47
Eric WofseyEric Wofsey
188k14216346
answered Jan 20 at 18:47
Eric WofseyEric Wofsey
188k14216346
answered Jan 20 at 18:47
Eric WofseyEric Wofsey
188k14216346
answered Jan 20 at 18:47
Eric WofseyEric Wofsey
188k14216346
188k14216346
$begingroup$
Thank you, but I keep reading that my quoted statement is the dual of Baer's criterion for injective modules and that this reduces to the Whitehead problem for $R= mathbb{Z}$. Do you have any idea what the correct version of this claim is supposed to look like?
$endgroup$
– CuriousKid7
Jan 20 at 19:05
$begingroup$
Where are you reading that? I don't think there's any sense in which the Whitehead problem is literally dual to Baer's criterion, though they are certainly related.
$endgroup$
– Eric Wofsey
Jan 20 at 19:35
$begingroup$
I've given the link of one reference in my OP. But you are certainly right.
$endgroup$
– CuriousKid7
Jan 20 at 19:45
2
$begingroup$
I think it's simply an error in that text where the authors were a bit over-eager to fit a remark on Whitehead's problem into their narrative.
$endgroup$
– Eric Wofsey
Jan 20 at 19:50
$begingroup$
Yes, I agree. In fact, the correct formulation is much more delicate. It seems to be known as "Faith's problem on $R$-projectivity" if you're interested.
$endgroup$
– CuriousKid7
Jan 20 at 19:52
add a comment |
$begingroup$
Thank you, but I keep reading that my quoted statement is the dual of Baer's criterion for injective modules and that this reduces to the Whitehead problem for $R= mathbb{Z}$. Do you have any idea what the correct version of this claim is supposed to look like?
$endgroup$
– CuriousKid7
Jan 20 at 19:05
$begingroup$
Where are you reading that? I don't think there's any sense in which the Whitehead problem is literally dual to Baer's criterion, though they are certainly related.
$endgroup$
– Eric Wofsey
Jan 20 at 19:35
$begingroup$
I've given the link of one reference in my OP. But you are certainly right.
$endgroup$
– CuriousKid7
Jan 20 at 19:45
2
$begingroup$
I think it's simply an error in that text where the authors were a bit over-eager to fit a remark on Whitehead's problem into their narrative.
$endgroup$
– Eric Wofsey
Jan 20 at 19:50
$begingroup$
Yes, I agree. In fact, the correct formulation is much more delicate. It seems to be known as "Faith's problem on $R$-projectivity" if you're interested.
$endgroup$
– CuriousKid7
Jan 20 at 19:52
$begingroup$
Thank you, but I keep reading that my quoted statement is the dual of Baer's criterion for injective modules and that this reduces to the Whitehead problem for $R= mathbb{Z}$. Do you have any idea what the correct version of this claim is supposed to look like?
$endgroup$
– CuriousKid7
Jan 20 at 19:05
$begingroup$
Thank you, but I keep reading that my quoted statement is the dual of Baer's criterion for injective modules and that this reduces to the Whitehead problem for $R= mathbb{Z}$. Do you have any idea what the correct version of this claim is supposed to look like?
$endgroup$
– CuriousKid7
Jan 20 at 19:05
$begingroup$
Where are you reading that? I don't think there's any sense in which the Whitehead problem is literally dual to Baer's criterion, though they are certainly related.
$endgroup$
– Eric Wofsey
Jan 20 at 19:35
$begingroup$
Where are you reading that? I don't think there's any sense in which the Whitehead problem is literally dual to Baer's criterion, though they are certainly related.
$endgroup$
– Eric Wofsey
Jan 20 at 19:35
$begingroup$
I've given the link of one reference in my OP. But you are certainly right.
$endgroup$
– CuriousKid7
Jan 20 at 19:45
$begingroup$
I've given the link of one reference in my OP. But you are certainly right.
$endgroup$
– CuriousKid7
Jan 20 at 19:45
2
2
$begingroup$
I think it's simply an error in that text where the authors were a bit over-eager to fit a remark on Whitehead's problem into their narrative.
$endgroup$
– Eric Wofsey
Jan 20 at 19:50
$begingroup$
I think it's simply an error in that text where the authors were a bit over-eager to fit a remark on Whitehead's problem into their narrative.
$endgroup$
– Eric Wofsey
Jan 20 at 19:50
$begingroup$
Yes, I agree. In fact, the correct formulation is much more delicate. It seems to be known as "Faith's problem on $R$-projectivity" if you're interested.
$endgroup$
– CuriousKid7
Jan 20 at 19:52
$begingroup$
Yes, I agree. In fact, the correct formulation is much more delicate. It seems to be known as "Faith's problem on $R$-projectivity" if you're interested.
$endgroup$
– CuriousKid7
Jan 20 at 19:52
add a comment |
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